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[Edited to give a complete answer]

To simplify notation, let me write $U_i$ for $V\smallsetminus W_i$, and $U_{12}$ for $U_1\cap U_2=V\smallsetminus(W_1\cup W_2)$.

Fact: The obvious functor $$(\mathrm{Sch}/V)\longrightarrow (\mathrm{Sch}/U_1) \times_{(\mathrm{Sch}/U_{12})} (\mathrm{Sch}/U_2)$$ is an equivalence. In other words, a $V$-scheme $X$ is the same thing as a $U_1$-scheme $X_1$, a $U_2$-scheme $X_2$, and a $U_{12}$-isomorphism of their restrictions to $U_{12}$.
This is probably somewhere in EGA1. [EDIT: all I could find was section 2.4 of EGA1, relying on (4.1.7) of Chapter 0 (glueing of riged spaces).]
However, we are dealing here with finite étale schemes, which happen to be affine over the base, so this boils down to the analogous statement for categories of quasicoherent sheaves, which is essentially trivial (plus the fact that finite étale'' is a local condition).

If we describe the categories of finite étale covers in terms of $\pi_1$-sets, the above equivalence says that the diagram of groups $$(*)\qquad\begin{array}{rcl} \pi_1(U_{12},p)=:G_{12}& \longrightarrow &G_1:=\pi_1(U_{1},p)\cr \downarrow && \downarrow\cr \pi_1(U_{2},p)=:G_2& \longrightarrow &G:=\pi_1(V,p) \end{array}$$ is cocartesian. In other words, we get the usual van Kampen statement: the natural map $$\pi_1(U_{1},p)\ast_{\pi_1(U_{12},p)}\pi_1(U_{2},p)\longrightarrow \pi_1(V,p)$$ is an isomorphism. [EDIT: the coproduct is in the profinite category, which perhaps makes it hard to describe in general.general. See Will Savin's comment.]

What we want to prove is that the map "on the other side" $$G_{12}\longrightarrow G_1 \times_G G_2$$ is surjective, given that all the maps in diagram ($\ast$) are surjective.

Identifying $G_i$ ($i=1,2$) with $G_{12}/N_i$, we see from the universal property of the coproduct that $G=G_{12}/N_{1}N_{2}$. [EDIT: clearly this works also in the profinite category: since $N_1$, $N_2$ are both compact normal subgroups, so is $N_1 N_2$, hence $G_{12}/N_{1}N_{2}$ is profinite].

Take any $(x_1,x_2)\in G_1 \times_G G_2$: thus we have $x_i=g_i N_i$ for some $g_i\in G_{12}$, and the fiber product condition says that $g_1=g_2 n_2 n_1$ for some $n_i\in N_i$ (recall that $N_1 N_2=N_2 N_1$). So, $(x_1,x_2)$ is the image of $g_1 n_{1}^{-1}=g_2 n_2\in G_{12}$. QED

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[Edited to give a complete answer]

To simplify notation, let me write $U_i$ for $V\smallsetminus W_i$, and $U_{12}$ for $U_1\cap U_2=V\smallsetminus(W_1\cup W_2)$.

Fact: The obvious functor $$(\mathrm{Sch}/V)\longrightarrow (\mathrm{Sch}/U_1) \times_{(\mathrm{Sch}/U_{12})} (\mathrm{Sch}/U_2)$$ is an equivalence. In other words, a $V$-scheme $X$ is the same thing as a $U_1$-scheme $X_1$, a $U_2$-scheme $X_2$, and a $U_{12}$-isomorphism of their restrictions to $U_{12}$.
This is probably somewhere in EGA1. However, we are dealing here with finite étale schemes, which happen to be affine over the base, so this boils down to the analogous statement for categories of quasicoherent sheaves, which is essentially trivial (plus the fact that finite étale'' is a local condition).

If we describe the categories of finite étale covers in terms of $\pi_1$-sets, the above equivalence says that the diagram of groups $$(*)\qquad\begin{array}{rcl} \pi_1(U_{12},p)=:G_{12}& \longrightarrow &G_1:=\pi_1(U_{1},p)\cr \downarrow && \downarrow\cr \pi_1(U_{2},p)=:G_2& \longrightarrow &G:=\pi_1(V,p) \end{array}$$ is cocartesian. In other words, we get the usual van Kampen statement: the natural map $$\pi_1(U_{1},p)\ast_{\pi_1(U_{12},p)}\pi_1(U_{2},p)\longrightarrow \pi_1(V,p)$$ is an isomorphism. [EDIT: the coproduct is in the profinite category, which perhaps makes it hard to describe in general.]

What we want to prove is that the map "on the other side" $$G_{12}\longrightarrow G_1 \times_G G_2$$ is surjective, given that all the maps in diagram ($\ast$) are surjective.

Identifying $G_i$ ($i=1,2$) with $G_{12}/N_i$, we see from the universal property of the coproduct that $G=G_{12}/N_{1}N_{2}$. [EDIT: clearly this works also in the profinite category: since $N_1$, $N_2$ are both compact normal subgroups, so is $N_1 N_2$, hence $G_{12}/N_{1}N_{2}$ is profinite].

Take any $(x_1,x_2)\in G_1 \times_G G_2$: thus we have $x_i=g_i N_i$ for some $g_i\in G_{12}$, and the fiber product condition says that $g_1=g_2 n_2 n_1$ for some $n_i\in N_i$ (recall that $N_1 N_2=N_2 N_1$). So, $(x_1,x_2)$ is the image of $g_1 n_{1}^{-1}=g_2 n_2\in G_{12}$. QED

These are some thoughts about what the question really is (in particular the relation

[Edited to van Kampen's theorem), not an give a complete answerto the question: ]

\pi_1(U_{12},p)pi_1(U_{12},p)=:G_{12}& \longrightarrow &\pi_1(U_{1},p)\cr\pi_1(U_{2},p)pi_1(U_{2},p)=:G_2& \longrightarrow &\pi_1(V,p)is cocartesian (not cartesian); in . In other words, we get the usual van Kampen statement: the natural map

However, I don't know what can be deduced from this in therms of

What we want to prove is that the map "on the other side" $$\pi_1(U_{12},p)\longrightarrow G_{12}\longrightarrow G_1 \pi_1(U_{1},p)\times_{\pi_1(V,p)}\pi_1(U_{2},p).$$If we have a times_G G_2is surjective, given that all the maps in diagram ($\ast$) are surjective.

Identifying $G_i$ ($i=1,2$) with $G_{12}/N_i$, we see from the universal property of groups the coproduct that $G_1\rightarrow G_3\leftarrow G=G_{12}/N_{1}N_{2}$. Take any $(x_1,x_2)\in G_1 \times_G G_2$: thus we have $x_i=g_i N_i$ for some $g_i\in G_{12}$, I don't know if there is a simple description of and the category of fiber product condition says that $(G_1\times_{G_3} G_2)$-sets in terms of g_1=g_2 n_2 n_1$for some$G_i$-sets n_i\in N_i$ ($1\leq i\leq 3$)recall that $N_1 N_2=N_2 N_1$). So, $(x_1,x_2)$ is the image of $g_1 n_{1}^{-1}=g_2 n_2\in G_{12}$. QED

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